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Chapter 5: Q57E (page 191)
The uncertainty in a particle's momentum in an infinite well in the general case of arbitrary is given by .
For the uncertainty vanishes but is perfectly finite for all other values. Thus, so long as is large enough (it is), the uncertainty principle is perfectly satisfied for all .
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We say that the ground state for the particle in a box has nonzero energy. What goes wrong with in equation 5.16 if n = 0 ?
If a particle in a stationary state is bound, the expectation value of its momentum must be 0.
(a). In words, why?
(b) Prove it.
Starting from the general expression(5-31) with in the place of , integrate by parts, then argue that the result is identically 0. Be careful that your argument is somehow based on the particle being bound: a free particle certainly may have a non zero momentum. (Note: Without loss of generality, may be chosen to be real.)
A study of classical waves tells us that a standing wave can be expressed as a sum of two travelling waves. Quantum-Mechanical travelling waves, discussed in Chapter 4, is of the form . Show that the infinite well鈥檚 standing wave function can be expressed as a sum of two traveling waves.
A finite well always has at least one bound state. Why does the argument of Exercises fail in the case of a finite well?
Consider the delta well potential energy:
Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E0.
(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by
(b) Sketch and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?
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